Question

what is the angle of projection for a bullet fired over a horizontal plane , when horizontal range is equal to 4 times the maximum height ​

Answer 1

Answer:

We ignore the trivial case where the angle of projection equals zero degrees:

 h=

2g

v

i

2

​

sin

2

θ

i

​

​

;R=

g

v

i

2

​

(sin2θ)

​

;3h=R

so,  

2g

3v

i

2

​

sin

2

θ

i

​

​

=

g

v

i

2

​

(sin2θ

i

​

)

​

or,  

3

2

​

=

sin2θ

i

​

sin

2

θ

i

​

​

=

2

tanθ

i

​

​

thus, θ

i

​

=tan

−1

(

3

4

​

)=53.1

0